# Exercise with Square Spiral

### Source

Hubert Shutrick has kindly communicated to me his investigation of a hand-drawn spiral, as bellow:

This is a simple construction that leads to a self-similar shape, concerning which several questions may be meaningfully asked as to the lengths and areas involved. Answers to these questions reveal that, the self-similarity notwithstanding, neither the boundary curves, nor the enclosed form, are fractals - just plain self-similar figures. The curves have finite lengths, whereas the enclosed shape has finite area. There are certainly weirder cases.

### Questions

There are several questions that may be asked concerning the resulting fractal. The shape consists of quarter circles stuck together. It is bounded by two spiral curves: the smooth outer spiral and the rectilinear inner spiral.

1. What is the length of the smooth outer spiral?
2. What is the length of the rectilinear inner spiral?
3. What is the area enclosed between the two curves?
4. What is the accumulation point of the two spirals?

For all questions the answer comes from a formula for a sum of the geometric series:

$\displaystyle a+aq+aq^2+aq^3+\ldots =a\frac{1}{1-q}=\frac{a}{1-q}.$

1. What is the length of the smooth outer spiral?

$\displaystyle a=\frac{2\pi}{4}=\frac{\pi}{2},\,$ $\displaystyle q=\frac{1}{2},\,$ $\displaystyle\frac{a}{1-q}=\pi.$

2. What is the length of the rectilinear inner spiral?

$a=1,\,$ $\displaystyle q=\frac{1}{2},\,$ $\displaystyle\frac{a}{1-q}=2.$

3. What is the area enclosed between the two curves?

$\displaystyle a=\frac{\pi}{4},\,$ $\displaystyle q=\frac{1}{4},\,$ $\displaystyle\frac{a}{1-q}=\frac{\pi}{3}.$

4. What is the accumulation point of the two spirals?

If the accumulation point - the tip of the spiral shape - is described as $(x_0,y_0),\,$ we'll compute the two separately:

$x_0=\displaystyle\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\ldots,\;$ so that $a=\displaystyle\frac{1}{4},\,$ $q=\displaystyle -\frac{1}{4}.\;$ Thus,

$\displaystyle\frac{a}{1-q}=\frac{1}{4}\cdot\frac{4}{5}=\frac{1}{5}.$

For $y_0,\,$ the calculations are the same, except $\displaystyle a=\frac{1}{2},\,$ so that $\displaystyle\frac{a}{1-q}=\frac{1}{2}\cdot\frac{4}{5}=\frac{2}{5}.$

We come up with the accumulation point $\displaystyle (x_0,y_0)=\left(\frac{1}{5},\frac{2}{5}\right).$

Hubert Shutrick's original solutions dependent on the shape at hend being self-similar. The first piece aside, the rest of the shape is constructed exactly like the whole but strating with half as large quarter circle (and this rotated $90^{\circ}\,$ degrees clockwise). This remark leads to a few simple equations:

1. What is the length of the smooth outer spiral?

If the sought length is $L\,$ then $\displaystyle \frac{\pi}{2}+\frac{L}{2}=L,\,$ implying $L=\pi.$

2. What is the length of the rectilinear inner spiral?

If the sought length is $M\,$ then $\displaystyle 1+\frac{M}{2}=M,\,$ implying $M=2.$

3. What is the area enclosed between the two curves?

If $S\,$ is the said area, $\displaystyle S=\frac{\pi}{4}+\frac{1}{4}S,\,$ implying $S=\displaystyle\frac{\pi}{3}.$

4. What is the accumulation point of the two spirals?

If $z\,$ is the accumulation we have $\displaystyle z=\frac{i}{2}-\frac{i}{2}z,\,$ implying

\displaystyle\begin{align} 2z&=i-iz,\\ (2+i)z&=i,\\ z&=\frac{i}{2+i}=\frac{i(2-i)}{5}=\frac{1}{5}+i\frac{2}{5}. \end{align}

It's possible to avoid complex variables by observing that $\displaystyle\left(\frac{3}{16},\frac{3}{8}\right),$ is the new origin of a copy size $\displaystyle\frac{1}{16}-\text{th}\,$ so the vector $v\,$ from the origin to the accumulation point satisfies

$\displaystyle v = \left(\frac{3}{16},\frac{3}{8}\right) + \frac{v}{16}.$

Finally, Hubert also observed that the homothety from the whole elephant to the one $\displaystyle\frac{1}{16}-\text{th}\,$ its size is a projection from the accumulation point which can therefore be also found as the intersection of the line of centres $y=2x\,$ and the line of starting points of the spirals $3y-x=1.$